Answer:
The Base is 150 and the Exponent is 4
Area accomodate just the surface of a figure ( Length and width ) while Volume accomodate
the surface and edge of a figure ( length and witdth and height )
If you have a graphing calculator (such as a TI-84), you can use the normalcdf feature by clicking on the blue "2nd" button, then the "vars" button and then choice 2. Since you are finding the proportion of hybrids that get over 61 mpg, the lower bound is 61, the upper bound is infinity (you can type in 99999), the mean is 57, and the standard deviation is 3.5. So... normalcdf(61,99999,57,3.5) = .1265. This means that 12.65% of the hybrids get over 61 mpg.
Answer:
Using z score formula:
X = z ∂ + µ
= 157.833
Step-by-step explanation:
Solution:
Mean = µ = 1262
Standard deviation = ∂ = 117
(a) 28th percentile for the number of chocolate chip.
P( z < z) = 28%
= 0.28
P( z<- 0.58) = 0.28
Z = -0.58
By using z score formula:
Z = x - µ /∂
-0.58= x – 117 / 1262
X = (- 0.58)(117) + (1262)
= 1194.14
(b) Middle 97% of bag.
P(-z < z < z) = 97%
= 0.97
P( z < z) – p(z < -z) = 0.97
2p(z < z) -1 = 0.97
2p (z < z) = 1 + 0.97
P(z < z) = 1.97 / 2
= 0.99
P(z < 2.33) = 0.99
Z ± 2.33
By using z score formula:
Z = x - µ / ∂
X = z ∂ + µ
= - 2.33 x 117 + 1262
=989.39
Z = 2.33
X = z ∂ + µ
= 2.33 x 117 + 1262
=1533.61
(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate.
By using standard normal table,
The z dist’n formula:
P(z < z ) = 25%
=0.25
P(z < -0.6745) = 0.25
Z = 0.6745
Using z score formula:
X = z∂ + µ
= - 0.6745 x 117 + 1262
= 1183.0835
First quartile = Q1 =1183.0835
The third quartile is:
P(z<z) = 75%
= 0.75
P(z < 0.6745) = 0.75
Z = 0.6745
Using z score formula:
X = z ∂ + µ
= 0.6745 x 117 + 1262
= 1340.9165
IQR = Q3 – Q1
= 1340.9165 – 1183.0835
= 157.833
